JEE Main 2020 (Online) 4th September Morning ShiftQuestion 1
If $(a+\sqrt2b\cos x)(a-\sqrt2b\cos y)=a^2-b^2$, where $a>b>0$, then $\frac{dx}{dy}$ at $\left(\frac\pi4,\frac\pi4\right)$ is :
Differentiation and its Application
Solution
Concept. Differentiate with respect to $y$ (treating $x$ as a function of $y$), then put $x=y=\frac\pi4$. Formulas used. (1) $\sqrt2\,b\sin\frac\pi4=\sqrt2\,b\cos\frac\pi4=b$Step 1 - Differentiate. $$-\sqrt2\,b\sin x\,\frac{dx}{dy}\left(a-\sqrt2\,b\cos y\right)+\left(a+\sqrt2\,b\cos x\right)\sqrt2\,b\sin y=0.$$ Step 2 - At $\left(\frac\pi4,\frac\pi4\right)$. By (1), $$-b(a-b)\frac{dx}{dy}+(a+b)\,b=0.$$ Step 3 - Solve. $\dfrac{dx}{dy}=\dfrac{a+b}{a-b}$. Answer: (C)